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If the lines $x+y-1=0, k x+2 y+1=0$ and $4 x+2 k y+7$ $=0$ are concurrent, then $k=$
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Verified Answer
The correct answer is:
$\frac{-13}{2}$
$\left|\begin{array}{ccc}1 & 1 & -1 \\ k & 2 & 1 \\ 4 & 2 k & 7\end{array}\right|=0$
$\begin{aligned} & \Rightarrow \quad(14-2 k)-(7 k-4)-\left(2 k^2-8\right)=0 \\ & \Rightarrow \quad 2 k^2+9 k-26=0 \\ & \Rightarrow \quad 2 k^2+13 k-4 k-26=0 \\ & \Rightarrow \quad k(2 k+13)-2(2 k+13)=0 \\ & \Rightarrow \quad(2 k+13)(k-2)=0 \\ & \Rightarrow \quad k=2, \frac{-13}{2}\end{aligned}$
But at $k=2$, lines will be coincident
$\therefore \quad k=\frac{-13}{2}$.
$\begin{aligned} & \Rightarrow \quad(14-2 k)-(7 k-4)-\left(2 k^2-8\right)=0 \\ & \Rightarrow \quad 2 k^2+9 k-26=0 \\ & \Rightarrow \quad 2 k^2+13 k-4 k-26=0 \\ & \Rightarrow \quad k(2 k+13)-2(2 k+13)=0 \\ & \Rightarrow \quad(2 k+13)(k-2)=0 \\ & \Rightarrow \quad k=2, \frac{-13}{2}\end{aligned}$
But at $k=2$, lines will be coincident
$\therefore \quad k=\frac{-13}{2}$.
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